Fourier Concepts

ES3

© 2001 KEDMI Scientific Computing. All Rights Reserved. 

Square wave example:

V(t) = 4/ sin(t)

+ 4/3 sin(3t)

+ 4/5 sin(5t)

+ 4/7 sin(7t)

+ 4/9 sin(9t)

+ …

time

Any periodic waveform can be constructed by adding sine and cosine waves with frequencies that are integer multiples of the waveform’s frequency.

added frequency

total signal

…frequency

|v|fo

5fo7fo 9fo 11fo

3fo

Frequency spectrum of a square wave

Flute

3.34 3.342 3.344 3.346 3.348 3.35 3.352 3.354 3.356

x 104

-0.04

-0.03

-0.02

-0.01

0

0.01

0.02

0.03

0.04

0.05

Time/44100 (sec)

Am

plitu

de

Flute Sample

0 1000 2000 3000 4000 5000 6000 70000

100

200

300

400

500

600

700

800

900

1000

Frequency (Hz)

Am

plitu

de

Piano

2.82 2.84 2.86 2.88 2.9 2.92

x 104

-0.06

-0.04

-0.02

0

0.02

0.04

0.06

0.08

Time (x44100)

Am

plitu

de

Piano Sample

0 500 1000 1500

2

4

6

8

10

12x 10

4

Frequency (Hz)

Am

plitu

de

Frequency Spectrum of Piano

4, 5, 6

fo = 260 Hz

2fo = 520 Hz

3fo = 780 Hz

Acoustic Bass

0 100 200 300 400 500 600

2

4

6

8

10

12

x 105

Frequency (Hz)

Am

plitu

de

Frequency Spectrum of an Acoustic Bass

4.3 4.4 4.5 4.6 4.7 4.8

x 104

-0.15

-0.1

-0.05

0

0.05

0.1

0.15

Time (x44100)

Am

plitu

de

Acoustic Bass Sample

fo

2fofo

2fo

Hearing and the Eara simplified explanation

“unwound” cochlea (3.5cm)

eardrum

0 30,000Hair cell number

ampl

itud

e of

vib

rati

on

= sin(2200t) +0.2sin(25000t)

200Hz5kHz

1961 Nobel Prize, Georg von Bekesy

Hearing and the Ear (cont.)

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5-1.5

-1

-0.5

0

0.5

1

1.5

time (s)

ampl

itude

200Hz and 5kHz: Signal vs. Time

0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.01-1.5

-1

-0.5

0

0.5

1

1.5

time (s)

am

plit

ude

Close up view

0 1000 2000 3000 4000 5000 6000 7000 8000 9000 100000

100

200

300

400

500

600

700

800

900

frequency (Hz)

ampl

itude

Frequency Spectrum

FFT demonstration

• A microphone creates a voltage that is proportional to the pressure of a sound wave.

• This voltage is converted to a sequence of numbers that are stored in the computer’s memory

• The Fast Fourier Transform (FFT) is used to extract the amplitude of each sinusoid from the sound signal.

Filters

Filtersare usually represented as two-port networks

Two-portNetwork

Vin(t)+

Vout(t)-

+-

Input Port Output Port

Filtersare usually represented as two-port networks

Transfer Function:H()

VIN+VOUT

-

+-

Input Port Output Port

inin

outout

V

VH

)(

Transfer Function Example

3.34 3.342 3.344 3.346 3.348 3.35 3.352 3.354 3.356

x 104

-0.04

-0.03

-0.02

-0.01

0

0.01

0.02

0.03

0.04

0.05

Time/44100 (sec)

Am

plit

ude

Flute Sample

INPUT Signal, vin(t)

0 1000 2000 3000 4000 5000 6000 70000

100

200

300

400

500

600

700

800

900

1000

Frequency (Hz)

Am

plitu

de

FFT of the INPUT Signal: VIN()

Transfer Function Example

H()

1

Transfer Function ExampleThe Transfer Function, H()

…a low-pass filter

Transfer Function Example

0 1000 2000 3000 4000 5000 6000 70000

100

200

300

400

500

600

700

800

900

1000

Frequency (Hz)

Am

plitu

de

H()

1

At 530 Hz: VOUT = VIN x 1

At 1060 Hz: VOUT = VIN x 1

At 1590 Hz: VOUT = VIN x 0.25

At 2120 Hz: VOUT = VIN x 0.0

0 1000 2000 3000 4000 5000 6000 70000

100

200

300

400

500

600

700

800

900

1000

Frequency (Hz)

Am

plitu

de

Transfer Function ExampleFFT of the OUTPUT Signal: VOUT()

A real low-pass filterR

1 / jCVin Vout

)(tan)(

1)(

1|)(|

)(tan1)(

1

1

11

1

1

22

1

22

~~~~

RCH

RCH

RCRC

VRCj

V

CjR

CjVV inininout

The Transfer Functionof a low-pass filter

100

101

102

103

104

10-2

10-1

100

frequency (Hz)

|H(w

)|

R=1000C=1 uF

=1/RCis the cut-off frequency|H()|=1/√2

1000 rad/s = 159 Hz

The Transfer Functionof a low-pass filter

100

101

102

103

104

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

frequency (Hz)

phas

e of

H(w

) in

deg

rees

45o phase shift

@ = 1/RC